Integration of a natural log

1. Aug 24, 2004

krusty the clown

[SOLVED] Integration of a natural log

I am asked to Integrate by parts

$$\int \ln(2x+1) dx$$

So,
$$\mbox{u}=ln(2x+1)$$
$$\mbox{du}=\frac{2}{2x+1}$$
$$\mbox{dv}=\mbox{dx}$$
$$\mbox{v}=\mbox{x}$$
I plug all of that in and I get,

$${\int \ln(2x+1) dx\}={\mbox{x}\ln(2x+1)}-{\int \frac{2x}{2x+1} \mbox{dx}}$$

At this point I tried doing another substitution letting u equal 2x, but this just brought me back to the original question.
After this anything I try to do doesn't lead me anywhere. I know it can be solved because the answer is in the back of the book.
First, is what I have done so far correct?
Second, how do I finish the problem?

Erik

2. Aug 24, 2004

jamesrc

Try approaching it like this:

Let θ = 2x+1, so that your integral is:

$$\int{\ln\left(2x+1\right) dx} = 2\int{\ln\theta d\theta}$$

Now you can integrate by parts like you did before, with:

u = ln(θ) --> du = dθ/θ
dv = dθ --> v = θ

You should get:

$$2\left[ \theta\ln\theta - \theta \right]$$

as your result. Now all you have to do is plug θ = 2x+1 back in to find the final indefinite integral.

3. Aug 24, 2004

HallsofIvy

Staff Emeritus
jamesrc's suggestion is good. If you want to go ahead do it the way you were, you can actually do the division $\frac{2x}{2x+1}= 1- \frac{1}{2x+1}$. Of course, to do the integral of $\frac{1}{2x+1}$, make the substition u= 2x+1 (which gets you right back to jamesrc's suggestion).

Last edited: Aug 25, 2004
4. Aug 25, 2004

krusty the clown

thanks for your help, I hate it when the answer is so obvious and I can't figure it out.

Thanks, Erik

5. Aug 25, 2004

krusty the clown

One final question, when you substituted theda, shouldn't the constant in front of the integral be 1/2 instead of 2???

u=2x-1
du=2dx
dx=(1/2)du

6. Aug 25, 2004

e(ho0n3

Yes, it should be 1/2.

7. Aug 25, 2004

jamesrc

Yes, sorry about that. At least that proves that you know what you're doing.